Question 86797

Probability-and-statistics/86797 (2007-06-28 17:19:31): Hey! I'm new to this whole thing and I am LOST. . I have been out of high school now for 8 years and I am just not grasping some of this. I have a few questions and I hope you can help.. 
Here are my choices. Please let me know if I am correct. 
1. D
2. B
3. C
4. C
5. I have no idea where to begin
6. I have no idea where to begin 

Questions 
The numbers 1 to 10 are written on sheets of paper and placed in a hat.
If one sheet of paper is selected at random then find the following 
***The probability the number is greater than 3
A. 1/10 
B. 3/10 
C. 8/10 
D. 7/10 I say answer D is correct.
D is the correct answer.
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***The odds the number is a 9
P(it is a 9)= 1/10; P(is not a 9)=9/10; odds for = [1/10]/[9/10= 1:9
A is the correct answer
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A. 1:9 
B. 1:10 I think answer B is correct
C. 1:11 
D. 1:1
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***The probability the number is odd OR greater than 3
P(odd or >3) = P(odd)+p(>3)-P(odd and >3)
= 5/10 + 7/10 - 3/10 =9/10
A is the correct answer
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A. 9/10 
B. 7/20 
C. 12/10 I say C
D. 3/10 
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***The probability the number is odd AND less than 5
P(odd AND <5) = P(odd)P(<5 | odd)= (1/2)*(2/5)= 1/5

A. 9/10 
B. 7/10 
C. 1/5 I say answer C
D. 4/5 
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The numbers 1 to 10 are written on sheets of paper and placed in a hat.
If two sheets are selected at random then find the following: 
***The probability both numbers are ODD with replacement.
Note: “with replacement” assumes the first selection is replaced back into the hat before the second selection is made.
A. 1/100 
B. 1/4. 
C. 2/9 
D. 1/5 
***The probability both numbers are ODD without replacement.
Note: “without replacement” assumes the first selection is NOT replaced back into the hat, but is held out, while the second selection is made.
Answer: P(odd and odd) =  = 5/10*4/9 = 20/90=2/9

A. 1/5 
B. 2/9 
C. 1/4 
D. 1/100
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***The probability the first number is ODD and the second number is EVEN,
without replacement.
P(odd and even) = P(odd)*P(even) = 5/10*5/9 = 25/90 = 5/18
A. 1/4 
B. 5/18 
C. 2/9 
D. 13/18 
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***The probability of selecting a 2 and then a 9, with replacement.
P(2 and 9) = 1/10*1*9 = 1/90

A. 19/90 
B. 1/5 
C. 1/90 
D. 1/100 
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***One card is selected from a deck of standard poker cards (assume a typical deck of 52 totals cards numbered from 2 thru 10 and Jack thru Ace) at random. Find the
probability the card is a spade OR less than 8 (that is, from 2 to 7 only).
P(spade OR <8)=P(spade) + P(<8) - P(spade and <8)
=13/52 + 24/52 - 6/52 = 31/52  
A. 31/52 
B. 37/52 
C. 39/338 
D. 3/26 
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***The State Pick-3 lottery uses 27 balls numbered 1 to 9 (3 each).
What is the probability that my ticket numbered “222” will win ?
Assume that the balls are NOT replaced after each selection. 
Answer: 3C3/27C3 = 1/2925
A. 1 / 19683 
B. 1 / 2925 
C. 2 / 6561 
D. 1 / 1

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Cheers,
Stan H.